Second partials of the Gaussian e^(-x^2-y^2)

Chain rule out, then product rule back in

The exponent is $u = -x^2 - y^2$. Holding $y$ fixed, $\frac{\partial u}{\partial x} = -2x$, so the first partials are the original bump scaled by the inner derivative.

Now $f_x$ is a genuine product of two things that both depend on $x$, so differentiating it in $x$ needs the product rule: the $-2x$ gives $-2$, and the exponential gives back $-2x$ times itself.

The mixed partial is easier, because differentiating $-2x\,e^{-x^2-y^2}$ in $y$ leaves the $-2x$ alone as a constant and only touches the exponential.

The mistake: skipping the product rule on the pure second partials

Writing $f_{xx} = 4x^2 e^{-x^2-y^2}$ is the usual error. It comes from applying the chain rule twice and never noticing that $-2x$ is a function of $x$ as well. The missing $-2$ is not a detail: it is the entire reason the surface curves downward near the origin, where $4x^2$ is small.

Test any candidate at $(0,0)$. The Gaussian has a peak there, so $f_{xx}(0,0)$ must be negative. The correct expression gives $-2$; the version without the product rule gives $0$ and would call the peak degenerate.

Note also that the mixed partial genuinely does not need the product rule, since $-2x$ is constant with respect to $y$. Knowing which of the three needs it is the whole skill here.

Reading the bump off the Hessian

Both first partials vanish only at the origin, since the exponential is never zero. There $f_{xx} = f_{yy} = -2$ and $f_{xy} = 0$, so $D = (-2)(-2) - 0 = 4 > 0$ with $f_{xx} < 0$: a local maximum, of height $f(0,0) = 1$.

The sign of $f_{xx}$ flips where $4x^2 - 2 = 0$, that is at $x = \pm 1/\sqrt{2} \approx 0.707$. Inside that band the cross section curves downward like the top of a hill, outside it curves upward as the tail flattens toward zero. That flip is what makes the Gaussian look like a bell rather than a paraboloid.

Adding the two pure second partials gives $\nabla^{2} f = (4x^{2}+4y^{2}-4)\,e^{-x^{2}-y^{2}}$, which vanishes on the unit circle. Inside that ring the average curvature is negative and the surface is dome shaped, outside it the average curvature is positive as the tails level off toward zero. So the single curvature $f_{xx}$ changes sign on a pair of straight lines, while the two curvatures added together change sign on a circle.

The mixed partial $4xy\,e^{-x^2-y^2}$ is zero along both axes and changes sign between quadrants, which says the surface is untwisted along the axes and twisted in between.